Proof: When , the theorem is evident. For , let the theorem be false, so that . Thus, is a non-empty set bounded above (for ). Therefore, by completeness axiom has the supremum. Let . Then . Thus, is also an upper bound of . But . This contradicts that . Hence, the assumption is false, and so the statement of the theorem is true.

Corollary 1: For any such that .
It follows from the theorem on replacing by and taking for .

Corollary 2: The set is bounded below but unbounded above.

Corollary 3: For any there exist such that .

Corollary 4: For any there exist such that .

Corollary 5: For any there exist unique such that .

Proof: By the completeness axiom the set being bounded above has the suprema, say . Thus, , i.e. , where . Since is a suprema and therefore it is unique.

The above integer is usually denoted by and is called the integral part of the number .

Corollary 6: For any there exist unique such that .

Corollary 7: For any there exist unique such that .

Example: For any there exist unique such that .

Solution: For real number by corollary 5, there exist unique such that

Or

Or

i.e.

The above example illustrates that every positive integer can be uniquely expressed as , for unique .

Such unique representation of natural members is sometimes very helpful in investigating the enumerability of certain sets.